(4%5E3x%2B1%2B8%5E2x-1-11)%3D0.jpeg "(x^2+4x-2)(4^3x+1+8^2x-1-11)=0")
Solving the Equation (x^2 + 4x - 2)(4^3x+1 + 8^2x-1 - 11) = 0
This equation involves a product of two expressions that equals zero. The key to solving this kind of equation is to utilize the Zero Product Property:
Zero Product Property: If the product of two or more factors is zero, then at least one of the factors must be zero.
Following this principle, we can break down the equation into two separate equations:
- x^2 + 4x - 2 = 0
- 4^3x+1 + 8^2x-1 - 11 = 0
Let's solve each equation individually:
Solving x^2 + 4x - 2 = 0
This is a quadratic equation, and we can solve it using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
Where a = 1, b = 4, and c = -2.
Plugging these values into the quadratic formula, we get:
x = (-4 ± √(4^2 - 4 * 1 * -2)) / 2 * 1 x = (-4 ± √(24)) / 2 x = (-4 ± 2√6) / 2 x = -2 ± √6
Therefore, the solutions to the first equation are x = -2 + √6 and x = -2 - √6.
Solving 4^3x+1 + 8^2x-1 - 11 = 0
This equation involves exponents. To simplify, let's make the following substitutions:
- y = 2^x
Now, the equation becomes:
- y^6 + y^4 - 11 = 0
This is another quadratic equation (but in terms of y). We can use the quadratic formula again, with a = 1, b = 1, and c = -11.
y = (-1 ± √(1^2 - 4 * 1 * -11)) / 2 * 1 y = (-1 ± √(45)) / 2 y = (-1 ± 3√5) / 2
We have two possible values for y. Now, we need to substitute back to find x:
-
y = (-1 + 3√5) / 2
- 2^x = (-1 + 3√5) / 2
- x = log₂((-1 + 3√5) / 2)
-
y = (-1 - 3√5) / 2
- 2^x = (-1 - 3√5) / 2
- x = log₂((-1 - 3√5) / 2)
Since the base of the logarithm is 2, and the expression inside the logarithm is negative, these solutions for x are complex numbers.
Conclusion
The solutions to the equation (x^2 + 4x - 2)(4^3x+1 + 8^2x-1 - 11) = 0 are:
- x = -2 + √6
- x = -2 - √6
- x = log₂((-1 + 3√5) / 2) (complex)
- x = log₂((-1 - 3√5) / 2) (complex)